Calculus help!

Test Preparation
1

<p>I’m doing this online Calculus AB course (CTY, actually) and they don’t really explain ANYTHING. Can someone explain how to do these three problems?</p>

<li>lim x–> infiniti: [ sqrt(9x + x^2) ] / [ x^4 - 7 ]</li>
</ol>

<p>For the next two, they want you to use their proof that lim x–>0: sin(x)/x = 1</p>

<p>24.* lim x–> pi/3: [ sin(x + pi/6) - 1 ] / [ x - pi/3 ]</p>

<li>lim x–> 0: [ csc(x) - 1 ] / [ x csc(x) ]</li>
</ol>

<p>*My friend said this can be solved using “ther other form of derivatives”, but there should be another way, right? Because we technically haven’t done derivatives yet (this is chapter 2 stuff :P).</p>

<p>Thanks for your help!</p>

2

<p>To shorten:
9.
for x--> infiniti:
lim [sqrt(9x+x^2)] / (x^4-7)=
lim {[sqrt(9x+x^2)]/x^4} / [(x^4-7)/x^4]=
lim {sqrt[(9x+x^2)/x^8]} / (1-7/x^4)=
lim [sqrt(9/x^7 +1/x^6)] / (1-7/x^4)=
[sqrt(0+0)] / (1-0) = 0</p>

<p>24.*
For x--> pi/3:
lim [sin(x+pi/6)-1] / (x-pi/3)=
lim [sin(x+pi/6)-sin(pi/2)] / (x-pi/3)=
lim [2cos((x+pi/6+pi/2)/2) sin((x+pi/6-pi/2)/2] / (x-pi/3)=
lim [2cos((x+2pi/3)/2) sin((x-pi/3)/2] / (x-pi/3)=
lim {[2cos((x+2pi/3)/2) sin((x-pi/3)/2]/2} / [((x-pi/3)/2)]=
lim [cos((x+2pi/3)/2)] * {[sin((x-pi/3)/2)] / [(x-pi/3)/2]}=
[cos((pi/3 +2pi/3)/2)] * [sin(t)/t], where t-->0
cos(pi/2) * 1 = 0 * 1 = 0</p>

<p>25.
lim is undefined.
Can it be a right answer ot there is a mistake in the expression?</p>

<p>It's possible I messed up with parentheses somewhere - just write it out on paper.</p>

<ul>
<li>
sin(a)-sin(b) = 2 cos((a+b)/2) sin((a-b)/2)</li>
</ul>

3

<p>Whoa they didn't teach that identity in the Precalc course... Thanks soo much!</p>

<p>I've tried graphing the equation. When I graph it as it is, I get no limit, but when I substitute 1 for x csc(x) (isn't that the same as x/sin(x), which is sin(x)/x inverted, which is equal to 1?), the limit appears to be infiniti. Is this an illegal operation?</p>

4

<p>It's been a long time since I worked on limits problems, but here goes:</p>

<ol>
<li>lim x--> 0: [ csc(x) - 1 ] / [ x csc(x) ]
= lim x--> 0: [ {1/sin(x)} - 1 ] / [ x {1/sin(x)} ]
= lim x--> 0: [ {1/sin(x)} - 1 ] * [ sin(x)/x ]
= lim x--> 0: [ {1/x} - {sin(x)/x} ]</li>
</ol>

<p>Looks like gcf101 is right, the limit is undefined here.</p>

5

<p>Thanks optimizederdad for verifying!</p>

<p>Theoneo - you are legal all right.</p>

<p>For x-->a:
if lim(f(x)) is defined and lim(f(x))not=0, then
lim(1/f(x)) = 1/[lim(f(x))].</p>

<p>When you graph, there is a vertical asymptote y=0.
f(x)-->(-inf.) for x-->(-0),
f(x)-->(+inf.) for x-->(+0).</p>

<p>With this laconic online course of yours you probably need a good Calc AB prep book. Were there any suggestions on this forum?</p>

6

<p>Well, CTY's online program allows you to enroll for periods of time, during which you can do all the man you can. I already passed Precalculus in time for to take my school's Precalc final and skip it. Now I'm just enriching myself for Calc next year :P.</p>

<p>I think I'm probably going to get Barron's or PR for the AP.</p>

7

<p>Maybe you could bum/borrow your school Calc book from somebody (your teacher?) who's done with it? That would get you ahead too.</p>

8

<p>Get Stewart! It has a great explanation of things like this.</p>

9

<p>what's Stewart?</p>

10

<p>Sorry, it is a calculus book by James Stewart.</p>